Rationalization

Summary: The technique of eliminating radicals from a denominator by multiplying numerator and denominator by a conjugate expression, introduced by Euler in the context of dividing irrational quantities.

Sources: chapter-1.2.8, chapter-2.0.4, chapter-2.0.8, chapter-2.0.9, chapter-2.0.10, additions-5

Last updated: 2026-05-10

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The Conjugate Trick

If the denominator is $a+\sqrt{b}$, multiplying top and bottom by $a-\sqrt{b}$ produces the rational denominator $a^2-b$ (§330): (source: chapter-1.2.8)

$$\frac{p}{a+\sqrt{b}}=\frac{p(a-\sqrt{b})}{a^2-b}.$$

The conjugate and original expression are a matched pair: their product is always rational (or integer) because the radical terms cancel.

Examples

From §330–331: (source: chapter-1.2.8)

Original	Multiply by	Result
$\frac{3+2\sqrt{2}}{1+\sqrt{2}}$	$1-\sqrt{2}$	$\sqrt{2}+1$
$\frac{8-5\sqrt{2}}{3-2\sqrt{2}}$	$3+2\sqrt{2}$	$4+\sqrt{2}$
$\frac{1}{5-2\sqrt{6}}$	$5+2\sqrt{6}$	$5+2\sqrt{6}$
$\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}$	$\sqrt{6}+\sqrt{5}$	$11+2\sqrt{30}$

Multi-Radical Denominators

When the denominator contains several distinct radicals, Euler eliminates them one at a time (§332). For example: (source: chapter-1.2.8)

$$\frac{1}{\sqrt{10}-\sqrt{2}-\sqrt{3}}\stackrel{\times (\sqrt{10}+\sqrt{2}+\sqrt{3})}{\to }\frac{\sqrt{10}+\sqrt{2}+\sqrt{3}}{5-2\sqrt{6}}\stackrel{\times (5+2\sqrt{6})}{\to }5\sqrt{10}+11\sqrt{2}+9\sqrt{3}+2\sqrt{60}.$$

Relation to Algebraic Identities

The conjugate technique rests on the difference-of-squares identity: (source: chapter-1.2.8)

$$(a+\sqrt{b})(a-\sqrt{b})=a^2-b.$$

The same identity from algebraic-identities that Euler derived in Chapter 1.2.3 is thus the algebraic engine behind denominator rationalization.

Rationalizing $\sqrt{a+bx+c{x}^2}$ (Part II, Chapter 4)

A deeper use of rationalization is to find rational values of $x$ making $a+bx+c{x}^2$ a perfect square, so that $\sqrt{a+bx+c{x}^2}$ becomes rational. Euler develops four rules (source: chapter-2.0.4):

Rule	Condition	Ansatz
1	$c=f^2$ is a square	$\sqrt{a+bx+f^2{x}^2}=fx+m/n$
2	$a=f^2$ is a square	$\sqrt{f^2+bx+c{x}^2}=f+mx/n$
3	$b^2-4ac=d^2$ is a square	set root $=m(f+gx)/n$ after factoring
4	Formula $=p^2+qr$ with $p,q,r$ linear	set root $=p+mq/n$

In each case the ansatz makes the square of the assumed root match the formula exactly, the highest-degree $x$ terms cancel, and a linear equation in $x$ remains — yielding an explicit rational solution depending on the free parameters $m$ and $n$.

See ch2.0.4-surd-rationalization for full derivations and worked examples.

Higher-Degree Radicands (Part II, Chapters 8–10)

The four-rule scheme extends — with significant qualifications — to cubic and quartic radicands and to cube-root extraction.

Chapter	Formula	Key restrictions
VIII	$\sqrt{a+bx+c{x}^2+d{x}^3}$	A seed solution is mandatory; each pass yields one new $x$; some formulas (e.g. $1+x^3$) admit only finitely many solutions
IX	$\sqrt{a+bx+c{x}^2+d{x}^3+e{x}^4}$	Three subclasses by which end is a square; up to six new values when both ends are squares; no method beyond degree 4 (§146)
X	$\sqrt[3]{a+bx+c{x}^2+d{x}^3}$	Three subclasses by which end is a cube; degree limit is 3 (one less than the squaring case)

The common ansatz pattern across all chapters: choose a polynomial root with enough free parameters to kill the leading or trailing terms, then solve the residual linear equation. Failure modes recur — formulas with only first and last terms (e.g. $f^2+e{x}^4$, $a+g^3{x}^3$) collapse to trivial $x$.

Frontier (Ch IX §146): degree 5 and higher cannot be transformed into squares by these techniques. Euler treats this as the limit of “the subject hitherto known.”

Lagrange’s Decision Procedure for $\sqrt{a+bx+c{x}^2}$ (Additions V)

Where Euler’s four rules of ch2.0.4-surd-rationalization are constructive but ad-hoc — each requires the radicand to fit a specific shape — Lagrange’s Add. V algorithm gives a uniform decision procedure:

1. Reduce $\sqrt{a+bx+c{x}^2}$ rational to making $A{y}^2+B$ a square (with $A=4c$, $B=b^2-4ac$, then strip square factors).
2. Equivalently, solve $A{p}^2+B{q}^2=z^2$ in coprime integers.
3. Apply the descent procedure: substitute $z=nq-A{q}^{\prime }$ with $|n|\le A/2$ and $A\mid n^2-B$, obtain a strictly smaller leading coefficient $A^{\prime }=(n^2-B)/A$, iterate.
4. Termination is guaranteed: either a square coefficient appears (success — back-substitute) or no valid residue exists (failure — proven impossible).

Once one rational $x=f$ with $a+bf+c{f}^2=g^2$ is found, all other rational solutions follow from the chord-method parametrization (Add. V Art. 57):

$x=\frac{f{m}^2-2gm+b+cf}{m^2-c},m\in \mathbb{Q}.$

This subsumes Euler’s four rules: each rule corresponds to a special seed $(f,g)$ that the radicand’s shape provides for free, and the parametrization sweeps out all other solutions.

Worked impossibility: $23y^2-5=\square$ has no rational solution — see add5-rational-quadratic-surds Art. 58 for the descent proof.

Related pages

• ch1.2.8-calculation-irrational-quantities
• ch2.0.4-surd-rationalization
• ch2.0.8-cubic-surd-rationalization
• ch2.0.9-quartic-surd-rationalization
• ch2.0.10-cubic-formula-as-cube
• algebraic-identities
• irrational-numbers
• square-roots-and-irrational-numbers
• imaginary-numbers
• indeterminate-analysis
• sum-of-two-cubes
• add5-rational-quadratic-surds
• lagrange-reduction-algorithm